NO3-
Na+ plus OH- plus H+ equals H2O plus Na+ plus Cl-
333 + 110 = 390 + 53
You would add 4.
x = 0 or x =-5
(0, 6)
The complete ionic equation for the reaction is: Ca^2+(aq) + 2NO3^-(aq) + 2K+(aq) + CO3^2-(aq) --> CaCO3(s) + 2K+(aq) + 2NO3^-(aq)
The net ionic equation for the reaction between KCl(aq) and Pb(NO3)2(aq) to form KNO3(aq) and PbCl2(s) is: 2K^+(aq) + 2NO3^-(aq) + Pb^2+(aq) + 2Cl^-(aq) -> 2K^+(aq) + 2NO3^-(aq) + PbCl2(s)
The spectator ions are NO3- in this reaction. They are present on both sides of the equation before and after the reaction takes place, so they do not participate in the reaction and can be considered spectators.
Na+ plus OH- plus H+ equals H2O plus Na+ plus Cl-
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I-
Ba2+(aq) + SO42-(aq) ---> BaSO4(S) Barium sulfate is insoluble in aqueous solution. Sodium nitrate is soluble so the sodium and nitrate ions are spectators that can be cancelled out from the ionic equation to give you this net ionic equation. Depending on the teacher, you may have to underline the products because it is a precipitate. (I posted this originally before signing up. Just a footnote).
The chemical equation is:Na + OH- + H+ + Cl- = Na+ + Cl- + H2O(l)
Molecular_equation_of_copper_II_sulfate_plus_sodium_carbonateCuSO4 + NaCO3 -------> Na2SO4 + CuCO3chebs
The chemical equation is:Na + OH- + H+ + Cl- = Na+ + Cl- + H2O(l)
The net ionic equation for the given reaction is H+ (aq) + OH- (aq) → H2O (l)
CuCl2(aq) + 2AgNO3(aq) = Cu(NO3)2(aq) + 2AgCl(s)
The balanced molecular equation is CaCl2 + Na2S -> CaS + 2NaCl. To write the ionic equation, we need to break down the reactants and products into their respective ions. This results in the ionic equation: Ca2+ + 2Cl- + 2Na+ + S2- -> CaS + 2Na+ + 2Cl-. Cross out spectator ions that appear on both sides of the equation to obtain the net ionic equation: Ca2+ + S2- -> CaS.